A remote control car is traveling in a straight line with a constant acceleration of \( 2 \mathrm{ft} / \mathrm{sec}^{2} \). If the initial position (at \( t=0 \) ) is 0 ft and the initial velocity is \( 1 \mathrm{ft} / \mathrm{sec} \), find the position at \( t=4 \) seconds. [ ? ] ft .

To find the position of the remote control car at \( t = 4 \) seconds, we can use the kinematic equation for position under constant acceleration: \[ s(t) = s_0 + v_0 t + \frac{1}{2} a t^2 \] Where: - \( s(t) \) is the position at time \( t \) - \( s_0 \) is the initial position (0 ft) - \( v_0 \) is the initial velocity (1 ft/sec) - \( a \) is the acceleration (2 ft/sec²) - \( t \) is the time (4 seconds) Plugging in the values: \[ s(4) = 0 + (1 \, \text{ft/sec}) \times (4 \, \text{s}) + \frac{1}{2} \times (2 \, \text{ft/sec}^2) \times (4 \, \text{s})^2 \] \[ s(4) = 0 + 4 \, \text{ft} + \frac{1}{2} \times 2 \times 16 \] \[ s(4) = 4 \, \text{ft} + 16 \, \text{ft} \] \[ s(4) = 20 \, \text{ft} \] Therefore, the position at \( t = 4 \) seconds is \( 20 \, \text{ft} \).